EDUCA ION b) Find the directional derivative \( D_{u} f(x, y, z) \) for \( f(x, y, z)=\sin (y z)+\ln x^{2} \) in the direction of the vector \( v=\langle 1,1,-1\rangle \). Hence find \( D_{u} f(1,1, \pi) \). [12 marks]

To find the directional derivative \( D_{u} f(x, y, z) \) of the function \( f(x, y, z) = \sin(yz) + \ln(x^2) \) in the direction of the vector \( v = \langle 1, 1, -1 \rangle \), we need to follow these steps: 1. **Normalize the direction vector** \( v \): \[ u = \frac{v}{\|v\|} = \frac{1}{\sqrt{3}} \langle 1, 1, -1 \rangle \] where \( \|v\| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \). 2. **Calculate the gradient** of \( f \): \[ \nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \] - \( \frac{\partial f}{\partial x} = \frac{2}{x} \) - \( \frac{\partial f}{\partial y} = z \cos(yz) \) - \( \frac{\partial f}{\partial z} = y \cos(yz) \) So, \[ \nabla f(x, y, z) = \left( \frac{2}{x}, z \cos(yz), y \cos(yz) \right) \] 3. **Evaluate the gradient at the point** \( (1, 1, \pi) \): \[ \nabla f(1, 1, \pi) = \left( 2, \pi \cos(\pi), 1 \cos(\pi) \right) = \left( 2, 0, -1 \right) \] 4. **Calculate the directional derivative**: \[ D_{u} f(1, 1, \pi) = \nabla f(1, 1, \pi) \cdot u \] This gives us: \[ D_{u} f(1, 1, \pi) = \left( 2, 0, -1 \right) \cdot \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) = \frac{2}{\sqrt{3}} + 0 + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \] So, the directional derivative \( D_{u} f(1, 1, \pi) \) is \( \sqrt{3} \).